• EE18
In summary, the conversation discusses the concept of constant variables in a thermodynamic process, specifically in an adiabatic process where heat is not exchanged. They talk about the first law of thermodynamics and how it relates to the change in internal energy of a system. They also mention the importance of considering extensive variables and homogeneous functions in order to determine the appropriate integration constants.
EE18
Homework Statement
Callen tells us the following:
Show that if a single-component system is such that ##PV^k## is constant in an
adiabatic process (k is a positive constant) the energy is given by
$$U = \frac{PV}{k-1} +Nf(\frac{PV^k}{N^k})$$
where ##f## is an arbitrary function.
Hint: ##PV^k## must be a function of S, so that ##\left( \frac{\partial U}{\partial V}
\right)_{S} = g(S)V^{-k}## where ##g(S)## is an unspecified function.
Relevant Equations
See my attempt below:
I was unable show that ##PV^k## must indeed equal some function of the entropy, ##g(S)##; maybe doing so would make things easier? I proceeded as below.
If we assume (as is almost surely intended by Callen) that in the given adiabatic (##d Q = 0##) process we are taking ##N## as constant and that the process is quasistatic, then we have both that ##dU = d W = -pdV## (from energy conservation/the first law) and that ##dU = TdS - pdV## (from the total differential of ##U## during this process). Thus ##TdS = 0## in such a process so that\footnote{I think we can take ##T \neq 0## as a reasonable assumption for any real process.} ##dS = 0## and thus ##S = const## during the process.

Now observe that if ##S## is held constant in a quasistatic process in which ##N## is constant, then we have ##dU = TdS - pdV +\mu dN = -pdV## while, also, ##dU = d Q -pdV## (first law for systems not changing ##N##) so that ##d Q = 0##. Thus by hypothesis, ##PV^k = const## in such a process. We'll call this constant ##C##, and note that in general it being "constant" means constant with respect to ##V##: ##C = C(S,N)## in general (why?).

Now we recall that
$$P := -\left( \frac{\partial U}{\partial V} \right)_{S,N};$$
thus, "integrating"\footnote{See the discussion \href{https://math.stackexchange.com/ques...r-a-function-given-a-particular-partial-deriv}{here} for what this means. \label{this}} the above leads to
$$U = -\int P dV = \frac{CV^{1-k}}{k-1} + h(S,N) \stackrel{(1)}{=} \frac{PV}{k-1} + h(S,N)$$
where $h$ is some other constant function.

I can't see how to go further than this though?

Edit: I've added the following argument but am not sure why ##c## below should be invertible?
Now we note that
##u = U/N## should be zeroth order homogeneous.Thus we have that (since each term of a first order homogeneous function -- here ##U## -- must itself be fst order homogeneous) ##H## must be first-order homogeneous. Thus we see that ##H(S,N) = NH(S/N,1) \equiv Nh(s).## Analyzing ##PV^k = C(S,N)## we see that since the left-side is ##k##th order homogeneous, so too must be the right hand side. So in particular ## C(S,N) = N^kC(S/N,1) \equiv N^kc(s)##. We now argue physically that ##c## should be invertible, so that ##s = c^{-1}(Pv^k)\$##so that, inserting this to the above gives
$$U = \frac{PV}{k-1} + Nh(s) = \frac{PV}{k-1} + Nh(c^{-1}(Pv^k)) \equiv \frac{PV}{k-1} + Nf(PV^k/N^k)$$
where ##f = g \circ c^{-1}##.

Last edited:
Are we allowed to bump?

EE18 said:
Are we allowed to bump?
Yes. Maybe @Chestermiller can help.

Hint: The "natural" independent thermodynamical variables for the internal energy, ##U##, for a one-component system are ##S##, ##V##, and ##N##, i.e., you have ##U=U(S,V,N)##.

Then you need the 1st Law
$$\mathrm{d} U=T \mathrm{d} S -P \mathrm{d} V +\mu \mathrm{d}N.$$
For an adiabatic change at constant ##N## you have given that
$$P V^k=C(S,N)=\text{const}.$$
Then you see from the 1st Law
$$\partial_V U(S,V,N)=-P.$$
This you can use to integrate the corresponding differential equation
$$-V^k \partial_V U=C(S,N).$$
Don't forget that the integration constant also depends on ##S## and ##N##.

Then you have to take into account that ##U##, ##S##, and ##V## are extensive variables, i.e., ##U## must be a homogeneous function of its variables of rank 1, i.e., you have
$$U(\lambda S, \lambda V,\lambda N)=\lambda U(S,V,N),$$

vanhees71 said:
Hint: The "natural" independent thermodynamical variables for the internal energy, ##U##, for a one-component system are ##S##, ##V##, and ##N##, i.e., you have ##U=U(S,V,N)##.

Then you need the 1st Law
$$\mathrm{d} U=T \mathrm{d} S -P \mathrm{d} V +\mu \mathrm{d}N.$$
For an adiabatic change at constant ##N## you have given that
$$P V^k=C(S,N)=\text{const}.$$
Then you see from the 1st Law
$$\partial_V U(S,V,N)=-P.$$
This you can use to integrate the corresponding differential equation
$$-V^k \partial_V U=C(S,N).$$
Don't forget that the integration constant also depends on ##S## and ##N##.

Then you have to take into account that ##U##, ##S##, and ##V## are extensive variables, i.e., ##U## must be a homogeneous function of its variables of rank 1, i.e., you have
$$U(\lambda S, \lambda V,\lambda N)=\lambda U(S,V,N),$$
Thank you so much for your answer! If it's possible to follow-up, I think everything up to "Don't forget that the integration constant also depends on ##S## and ##N##." gets us to
$$U = \frac{PV}{k-1} + H(S,N)$$ as I state in the OP. Admittedly, your method seems much more rigorous to arrive here.

For simplifying the arguments for the constant as you suggest in your last paragraph, does my edit in the OP correspond to what you had in mind?

vanhees71
Yes, that looks good!

EE18

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